Yi Dao high school physics questions, thank you

The first one should be the surface of the table, right?

The second should be the focus of the book.

This question is not very meaningful, it is just a question of statement, and it will definitely not be tested in the actual question, so it is good to understand it yourself.

The pressure of the book on the horizontal downward side should be the center of gravity of the book, and if it is the pressure of the table on the book, the action point should also be the center of gravity of the book.

As for why, generally speaking, if it only acts in the vertical direction, it is regarded as exerting force at a point in the center of gravity. This is especially true for you. Other shapes of objects are not known. This is what I think, so let's refer to it.

The downward pressure of the book on the horizontal plane will act on the contact surface of the horizontal plane.

If it is the pressure of the table on the book, the point of action is the book.

Why?

The surface of the first ten tables.

The second is the focus of the book.

The surface of the table.

The surface of the book!

are at the center of the book.

The rules are like this.

I think the point of action of the book on the downward pressure of the horizontal plane is on the horizontal plane.

The pressure of the table on the book (it should be elastic!) The point of action is on the book.

The points of action are all at the center of gravity of the book.

Because they are a pair of action and reaction forces, equal in magnitude and opposite in direction.

From the time the block falls in the river to the time it returns to catch up, the block and the boat depart from the same place and travel opposite each other, and the sum of the two speeds is "the speed at which the boat advances in still water"; From the return chase to the catch up with the block, the block and the boat are the pursuit problem, and the speed difference between the two is "the speed at which the boat advances in still water". Since the distance of the opposite side is the distance difference of = the pursuit of the problem, the time of the opposite = the time of the pursuit of the problem. So the block actually drifted for 2 hours, which included an hour before the chase and an hour before the chase.

I don't know if you know?

In fact, there is a simpler way, which is to use the flowing river water as a reference, then the wooden block is stationary, and the speed of the boat against the current and downstream is the speed in the still water. It is equivalent to the boat going back and forth from A (bridge: where the wooden block fell) to B (where the wooden block was found), and the speed and distance of the round trip are equal, and the time is 1 hour, and the wooden block floats for 2 hours.

Then the reference object was changed, and the water velocity was 1750m h. after drifting 3500 m in 2 hours

When the plank fell, the person did not notice it, but continued to move upstream, and only turned around when he found it. If the water velocity is x and the ship is still water velocity v, then the plank has drifted xt meters when it is found, and the person is moving forward (v-x)t. then the person and the plank are separated by vt.

Then an hour catches up, the plank advances x, and the man advances (v+x). If we look at vt=(v+x)-x, then t=1. That is, t=1.

That is, at the time of discovery, the ship had already advanced for 1 hour. So, the planks drifted for a total of 2 hours.

Taking the river water as the reference system, the wooden block is stationary after falling, the boat takes the lead to leave the wooden block at the same speed, and then moves closer to the wooden block, and the boat takes 1h to move closer to the wooden block, so it also takes 1h when the boat leaves the wooden block, and the time from the fall of the wooden block to the boat catching up with the wooden block is 2h.

h: hours).

1000m = a1t^2/2 + v2t + a2t^2/2 = t^2/2 + 40t - t^2/2 = 40t

t = 1000 Shi Wei 40 = 25s

After 25s, the front end of the two cars met to search for Sui Pei.

At this point v1 = 25m s v2 = 40 - 25 = 15m s200m = v1t + a1 2 2 + v2t + a2 2 2 = 25t + t 2 2 + 15t - t 2 2

40tt = 200/40 = 5s

After 5s, the two carts began to leave each other.

You don't have a question that says A is M or B is M.

Let's say A is called M.

Let the cross-sectional area be s, the length of m is 2x, the gravity is g1, the length of n is 2y, the gravity is g2, and the density of b(n) is

Easy to know g1=2xs*2 ; g2=2ys* then the root code chain lists the moment equilibrium equation according to the lever principle

g1*x=g2*y

i.e.: 4x 2s = 2y 2s

Solution, Chi Xun y = 2x

So the magnitude of the gravitational force of m is a multiple of n: g1 g2 = 2x y = 2

Isn't this a freshman physics question? It is important to note that each pumping does not take away the same gas even though it is the same volume, because the gas becomes thinner after each pumping. Each time the pump is full, the requirement is that the pressure in the cylinder is equal to the pressure in the container.

Considering any pumping process in the middle, when the pumping is full, the pumping cylinder has not been pulled out, at this time, because the pressure on both sides should be equal, in fact, the volume of gas becomes 11 10 times of the original, and the pressure becomes 10 11Obviously, the pressure before and after each pumping becomes the original 10 11After 10 times, it is (10 11) 10

Answer: There is a question with your answer. The difference is 10 times.

I don't know if you read the wrong answer.

Answer: The pressure of one draw becomes the original 10 11. Draw ten times to the tenth power of the original (10 11), it should be.

1. The uniform acceleration time is the longest. When the speed of F can reach 8000N, the speed has been uniform, but due to the increase of speed, the power used by the engine is also increasing, when the power reaches the rated power, the speed still increases, but it is not a uniform acceleration, that is, after the speed is greater than a certain value, due to the limitation of power, F can no longer maintain 8000N...

So, p = f max * v1, and the final velocity of the acceleration motion with acceleration of 2 v1 = 10m s, so, v = at, so, t = 5 seconds.

2. The distance problem under the rated power is considered to be done by the kinetic energy theorem.

When the car is driven at rated power for another 20 seconds, the engine workmanship w = pt = 80000 * 20 = 1600000j

The work done by the engine is converted into the increase of the kinetic energy of the car and the work done by overcoming the frictional force, that is, w=e increase + f=m(v2) 2 2-m(v1) 2 2+

The solution yields s=325m

（1）t=vmax/a=10 s

2) f = f total - f = f total.

F total = 8000 N, at this time v = at = 6m sp = f total * v = 48000 W

The maximum speed that can be achieved is V1, P=V1*F, V1=P F=12m SWtotal=P*T=960000 J=Fs

According to the principle of independence of motion synthesis and decomposition, both boxes in the vertical direction are in free fall! Due to the interval seconds, it is impossible for boxes A and B to be on the same level;

In the horizontal direction, because the two boxes are subjected to the same wind force, and the wind direction is opposite to the initial velocity direction, the two boxes move in a straight line at a uniform deceleration speed in the horizontal direction, and the acceleration of the deceleration is the same. Box A releases the horizontal initial velocity v first. The horizontal velocity of the rear A box is less than V.

Whereas, the horizontal velocity of box B is V. and box A is at the bottom of box B. After that, AB continues to decelerate with the same acceleration, and the horizontal velocity of A is always less than B, so the distance between ABs is getting bigger and bigger!

The answer is definitely c

I think the answer is more reasonable.

As can be seen from the title, when the object is released, it is affected by four forces, the vertical gravity is downward, and the air resistance is upward; Horizontally, both the wind and air resistance are to the left. Because the topic only discusses the horizontal direction, according to the independence of motion, so only consider the acceleration in the horizontal direction, from the question, it can be seen that the horizontal acceleration must be the opposite of the direction of velocity, so list the equation of motion (displacement - time), and then the subtraction of two shifts is a displacement relative to the other, you can see that the displacement of 2 objects is constantly decreasing relative to 1 object, I don't know if it is reasonable

Pick C, that's right.

Let the acceleration in the horizontal direction be -a and the velocity of the aircraft is v, then the velocity of the first box has become v when the second box is exactly leaving the plane'=, the horizontal movement distance of box two s2 = vt-1 2at 2, ( means square ha), the horizontal movement distance of box one s1 = v't-1 2at 2=(,(Of course, the box has a horizontal displacement at the beginning, but this does not affect the solution.) Then s2-s1= , will increase over time.

Flying against the wind, Box 1 slows down horizontally due to the wind, and Box 2 slows down further back.

At rest, up:down = 6:10

When exercising, up:down = 1:2 = 5:10

It can be seen that the above "lighter" = the "top force" of the spring on the top has become smaller = the box is doing upward acceleration telematic.

The idea of categorical discussion is very important.

Order: The mass of the object m, the gravitational force is downward, is g. The force exerted on the object by the upper top plate is downward, which is t=6. The force exerted by the spring on the object is upward, which is f=10.

At rest, t + g = f, 6 + g = 10, g = 4 (N), m = g g = 4 10 =

Hypothesis: If the object moves downward with acceleration, then t + g > f.

g does not change, f does not change (if the spring length does not change, f does not change) t becomes larger.

t f > 6 10 can never be.

The assumption is not true.

Hypothesis: If the object moves upwards at an accelerated pace, then t + g < f.

g does not change, f does not change (if the spring is shortened, the object will be removed from the upper surface, so that t = 0 and t f = 0.) So the spring will not be shortened)

t becomes smaller. t/f = ,t/10 =

The t=5 hypothesis is true.

The acceleration is AA = F(resultant force) M

f - t - g）/

meters per second squared).

To sum up: the object moves in a straight line with a uniform acceleration of acceleration of meters per second square, and the direction of acceleration is upward.

To solve this problem, first make it clear: the elastic force in the spring is always the same. The reason is that the sensors of the upper and lower top plates have always had an indicator, indicating that the metal block and spring have been on the upper and lower top plates, and indirectly indicating that the position of the metal block and spring has always been unchanged, so the length of the spring remains unchanged, and the elastic force remains the same.

And, since the upper top plate is only affected by the spring elastic force, it can be concluded that the spring elastic force is 6N.

Assuming that the box moves downward (or upwards decelerates), the block of metal has a downward acceleration. At this point, part (or all) of gravity produces an acceleration of the metal block (the so-called "weightlessness"), and as a result, the pressure on the lower roof plate decreases to less than 10n;The pressure of the upper roof has always been the spring elastic force, which is 6N, so it is impossible to say that "the indicator of the upper roof plate sensor is half of the number of the lower roof plate sensor", and the assumption is not valid.

Since it cannot be a downward acceleration, it must be an upward acceleration, and the corresponding force can only be generated by the lower roof plate. The pressure of the upper roof plate is always 6N, and it can be concluded that the pressure of the lower roof plate is 12N, and the extra 2N is the force corresponding to the acceleration. a=f m=2 is obtained from f=ma, that is, the box is doing upward uniform acceleration linear motion or downward uniform deceleration linear motion, and the acceleration is 5m s2.

Is there a picture??I don't quite understand the topic.